3
$\begingroup$

Given $K = \mathbb{Q}(t)$, $f(X) = X^5-t \in K[X]$ with $t$ trancedental over $\mathbb{Q}$.

a) Is $f$ irreducible?

b) Determine the degree of the splitting field of $f$ over $K$.

c) Determine the Galois group of $f$ over $K$.

My solutions so far:

a) I was thinking, $f$ is irreducible, since it is an eistenstein polynomial at $t$.

b) $\zeta^k\sqrt[5]{t}$ are the roots of $f$ for $k \in \{0,1,2,3,4\}$. Let $L$ the splitting field of $f$ over $K$. So $\sqrt[5]{t}$ is a element of the splitting field, and has minimal polynomial $X^5-t$. So $[\mathbb{Q}(\sqrt[5]{t}):\mathbb{Q}] = 5$. Now $\zeta$ has minimal polynomial $X^4+X^3+X^2+X+1$ which is degree 4. So $[\mathbb{Q}(\zeta):\mathbb{Q}] = 4$. But since $gcd(4,5) =1$ and $L = \mathbb{Q}(\zeta,\sqrt[5]{t})$. Then $[L:\mathbb{Q}] = 4\cdot 5 =20$. Is this correct reasoning?

c) Since $f$ has distinct roots, the given extension is a Galois extension and therefore $|Gal(L/K)| =20$.

I don't know how to continue, but found something in my book: Consider $\sigma(\sqrt[5]{t}) = \zeta\sqrt[5]{t}, \sigma(\zeta) = \zeta$, and consider $\tau(\sqrt[5]{t})=\sqrt[5]{t}, \tau(\zeta) = \zeta^2$. These have properties $ord(\sigma)=5, ord(\tau) =4$. But how to continue? or is it not possible to apply this method on this problem?

$\endgroup$
0
$\begingroup$

First just a short remark concerning b): You accidentally wrote $\mathbb{Q}$ in some places where it should have been $\mathbb{Q}(t)$ instead, for example, $[\mathbb{Q}(\sqrt[5]{t}):\mathbb{Q}(t)]=5$, and, accordingly, you have to show that $\phi_5(X)=X^4+X^3+X^2+X+1$ is indeed the minimal polynomial for $\zeta$ over $K=\mathbb{Q}(t)$.

Now for c). I'm not sure if this is what the the hint in the book aimed for, but here is a suggestion:

From the main theorem of Galois theory, we know that as $\mathbb{Q}(t)\subset \mathbb{Q}(t)(\zeta)$ is a normal extension (of degree $4$), the subgroup $\mathrm{Gal}(L/\mathbb{Q}(t)(\zeta))=:U\subset G:=\mathrm{Gal}(L/K)$ is normal (of index $4$, in particular, $\lvert U\rvert = 5$). Then, as $\lvert G\rvert = 20$ by b), $G$ must also have some subgroup $V$ of order $4$ (a $2$-Sylowsubgroup). As $\mathrm{gcd}(4,5)=1$, we have $U\cap V=\{\mathrm{Id}\}$, so we know that $G$ is a semidirect product, $$G\simeq U\rtimes_{\varphi}V,\quad\text{with }\begin{aligned}[t] \varphi\colon V&\rightarrow\mathrm{Aut}(U)\\v&\mapsto (u\mapsto vuv^{-1}).\end{aligned}$$
Regarding the hint of the book, $U=\langle \sigma\rangle $ and $V$ might be chosen as $V=\langle\tau\rangle$. The semidirect product is then completely determined by the relation established by $\varphi(\tau)$, i.e. by knowing the value of $\varphi(\tau)(\sigma)=\tau\sigma\tau^{-1}$: $$G=\langle\sigma,\tau\vert\mathrm{ord}(\sigma)=5,~\mathrm{ord}(\tau)=4,~\tau\sigma\tau^{-1}=\sigma^2\rangle$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.